reserve a,b,c for boolean object;
reserve p,q,r,s,A,B,C for Element of LTLB_WFF,
        F,G,X,Y for Subset of LTLB_WFF,
        i,j,k,n for Element of NAT,
        f,f1,f2,g for FinSequence of LTLB_WFF;
reserve M for LTLModel;

theorem Th36:
  for A be Element of LTL_axioms holds
  (A is LTL_TAUT_OF_PL or A is axltl1 or A is axltl1a or
  A is axltl2 or A is axltl3 or A is axltl4 or A is axltl5 or A is axltl6)
 proof
  defpred P1[Element of LTL_axioms] means
   $1 is LTL_TAUT_OF_PL or$1 is axltl1 or$1 is axltl1a or$1 is axltl2 or$1 is
axltl3 or$1 is axltl4 or$1 is axltl5 or$1 is axltl6;
  set X={p where p is Element of LTL_axioms:P1[p]};
  X c=LTLB_WFF
  proof
   let x be object;
   assume x in X;
   then ex p be Element of LTL_axioms st x=p & P1[p];
   hence x in LTLB_WFF;
  end;
  then reconsider X as Subset of LTLB_WFF;
  let A be Element of LTL_axioms;
  X is with_LTL_axioms
  proof
   let p,q;
   thus p is LTL_TAUT_OF_PL implies p in X
   proof
    assume A1: p is LTL_TAUT_OF_PL;
    then reconsider p1=p as Element of LTL_axioms by Def17;
    P1[p1] by A1;
    hence thesis;
   end;
   thus('not'('X' p))=>('X'('not' p)) in X
   proof
    reconsider pp=('not'('X' p))=>('X'('not' p)) as Element of LTL_axioms by
Def17;
    P1[pp] by Def22;
    hence thesis;
   end;
   thus('X'('not' p))=>('not'('X' p)) in X
   proof
    reconsider pp=('X'('not' p))=>('not'('X' p)) as Element of LTL_axioms by
Def17;
    P1[pp] by Def23;
    hence thesis;
   end;
   thus('X'(p=>q))=>(('X' p)=>('X' q)) in X
   proof
    reconsider pp=('X'(p=>q))=>(('X' p)=>('X' q)) as Element of LTL_axioms by
Def17;
    P1[pp] by Def24;
    hence thesis;
   end;
   thus('G' p)=>(p '&&'('X'('G' p))) in X
   proof
    reconsider pp=('G' p)=>(p '&&'('X'('G' p))) as Element of LTL_axioms by
Def17;
    P1[pp] by Def25;
    hence thesis;
   end;
   thus(p 'U' q)=>(('X' q)'or'('X'(p '&&'(p 'U' q)))) in X
   proof
    reconsider pp=(p 'U' q)=>(('X' q)'or'('X'(p '&&'(p 'U' q)))) as Element
of LTL_axioms by Def17;
    P1[pp] by Def26;
    hence thesis;
   end;
   thus(('X' q)'or'('X'(p '&&'(p 'U' q))))=>(p 'U' q) in X
   proof
    reconsider pp=(('X' q)'or'('X'(p '&&'(p 'U' q))))=>(p 'U' q) as Element
of LTL_axioms by Def17;
    P1[pp] by Def27;
    hence thesis;
   end;
   thus(p 'U' q)=>('X'('F' q)) in X
   proof
    reconsider pp=(p 'U' q)=>('X'('F' q)) as Element of LTL_axioms by Def17;
    P1[pp] by Def28;
    hence thesis;
   end;
  end;
  then A in LTL_axioms & LTL_axioms c=X by Def18;
  then A in X;
  then ex p be Element of LTL_axioms st A=p & P1[p];
  hence P1[A];
 end;
